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Energy gap in a p – n photodiode is 2.8 ...

Energy gap in a p – n photodiode is 2.8 eV. Can it detect a wavelength of 6000 nm? Justify your answer.

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Energy of photon E = `(hc)/(lambda) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(6000 xx 10^(-9) xx 1.6 xx 10^(19))eV = 2.06eV`
As `E lt E_(g) (2.8eV)`, so photodiode cannot detect this problem.
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